Aptitude
1. Principal: The money borrowed or lent out for a certain period is called the
principalor the sum.
2. Interest: Extra money paid for using other's money is called interest.
3. Simple Interest (S.I.) : If the interest on a sum borrowed for a certain period is reckoned uniformly, then it is called simple interest.
Let Principal = P, Rate = R% per annum (p.a.) and Time = T years. Then,
    i). S.I. =  (P*R*T)/100
   ii). P = (100*S.I)/(R*T)
       R = (100*S.I)/(P*T)
       T = (100*S.I)/(P*R)

1. Find the simple interest on Rs. 68,000 at 16 2/3% per annum for 9 months.

P = Rs.68000 
R = 50/3% p.a T = 9/12 years = 3/4 years S.I. = (P*T*R*)/100 = (68,000*(50/3)*(3/4)*(1/100)) = 8500 R.s

2. Find the simple interest on Rs. 3000 at 6 1/4% per annum for the period from 4th February 2005 to 18th April 2005.

Time = (24+31+18)days = 73 days = 73/365 years = 1/5 years.
P = Rs.3000
R = 6 ¼ % p.a = (25/4)% p.a
S.I = (3,000*(25/4)*(1/5)*(1/100)) = 37.50 R.s

Remark : The day on which money is deposited is not counted while the day on which money is withdrawn is counted .

3. A sum at simple interests at 13 ½ % per annum amounts to Rs.2502.50 after 4 years find the sum.

Let sum be Rs. x then , 
S.I. = Rs.(x*(27/2) *4*(1/100) ) = Rs.27x/50
amount = (Rs. x+(27x/50)) = Rs.77x/50
       =  77x/50 = 2502.50 => 77x = 2502.50 * 50 =>  x = 1625                                                          

4. A sum of Rs. 800 amounts to Rs. 920 in 8 years at simple intere interest rate is increased by 8%, it would amount to bow mucb ?

S.l. = Rs. (920 - 800) = Rs. 120
P = Rs. 800
T = 3 yrs. 
R = ((100 x 120)/(800*3)) % = 5 %.

New rate = (5 + 3)% = 8%.
New S.l. = (800*8*3)/100 = 192 Rs.
New amount = (800+192) = 992 Rs.

5. Adam borrowed some money at the rate of 6% p.a. for the first two years , at the rate of 9% p.a. for the next three years , and at the rate of 14% p.a. for the period beyond five years. 1£ he pays a total interest of Rs. 11, 400 at the end of nine years how much money did he borrow ?

Let the sum borrowed be x. Then,

(x*2*6)/100 + (x*9*3)/100 + (x*14*4)/100 = 11400
=> (3x/25 + 27x/100 + 14x / 25) = 11400       
=> 95x/100 = 11400 
=> x = (11400*100)/95 = 12000.
Hence, sum  borrowed = Rs.12,000.

6. A certain sum of money amounts to Rs. 1008 in 2 years and to Rs.1164 in 3 ½ years. Find the sum and rate of interests.

S.I. for 1 ½ years = Rs.(1164-1008) = Rs.156.
S.l. for 2 years = Rs.(156*(2/3)*2)=Rs.208         
Principal = Rs. (1008 - 208) = Rs. 800.
Now, P = 800, T = 2 and S.l. = 208.
Rate =(100* 208)/(800*2)% = 13%

7. At what rate percent per annum will a sum of money double in 16 years.

Let principal = P. 
Then, S.l. = P and T = 16 yrs.
Rate = (100 x P)/(P*16) % = 6 1/2 % p.a.         

The simple interest on a sum of money is 4/9 of the principal .Find the rate percent and time, if both are numerically equal.

Let sum = Rs. x. Then, S.l. = Rs. 4x/9
Let rate = R% and time = R years.
Then, (x*R*R)/100=4x/9 or R2 =400/9 or R = 20/3 = 6 2/3.
Rate = 6 2/3 %   and Time = 6 2/3 years = 6 years 8 months.

9. The simple interest on a certain sum of money for 2 l/2 years at 12% per annum is Rs. 40 less tban the simple interest on the same sum for 3 ½ years at 10% per annum. Find the sum.

Let the sum be Rs. x Then, ((x*10*7)/(100*2)) – ( (x*12*5)/(100*2)) = 40                     
                => (7x/20)-(3x/10) = 40      
                => x = (40 * 20) = 800.
Hence, the sum is Rs. 800.

10. A sum was put at simple interest at a certain rate for 3 years. Had it been put at 2% higher rate, it would have fetched Rs. 360 more. Find the sum.

Let sum = P and original rate = R.
Then, [ (P*(R+2)*3)/100] – [ (P*R*3)/100] = 360.
=> 3PR + 6P - 3PR = 36000 Û 6P=36000 Û P=6000
Hence, sum = Rs. 6000.

11. What annual instalment will discharge a debt of Rs. 1092 due in 3 years at 12% simple interest?

Let each Instalment be Rs. x
Then, ( x + ((x*12*1)/100) ) + ( x + ((x*12*2)/100) ) + x = 1092
=> ((28x/25) + (31x/25) + x) = 1092   
=> (28x + 31x + 25x) = (1092*25)
=> x = (1092*25)/84 = Rs.325.  
Each instalment = Rs. 325.

12. A sum of Rs. 1550 is lent out into two parts, one at 8% and another one at 6%. If the total annual income is Rs. 106, find the money lent at each rate.

Let the sum lent at 8% be Rs. x and that at 6% be Rs. (1550 - x).
((x*8*1)/100) + ((1550-x)*6*1)/100 = 106
=> 8x + 9300 – 6x = 10600 
=> 2x = 1300  
=> x = 650.
Money lent at 8% = Rs. 650. 
Money lent at 6% = Rs. (1550 - 650) = Rs. 900.

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